#### Answer

-5,2

#### Work Step by Step

Horizontal lines have slope equal to 0, so to determine the values of $x$ where the tangent line is horizontal, we set the derivative of $f$ equal to $0$ and solve for $x$.
Using the Power Rule, the derivative of $$f(x)= 2x^3 +9x^2-60x+4$$ is
$$f'(x) = 6x^2+18x-60.$$
We set the derivative $f'(x)$ equal to 0 and solve for $x$:
$$f'(x) = 6x^2+18x-60=0.$$
Factor out a 6 from the left side:
$$6( x^2+3x-10)=0.$$
Dividing both sides of the equation by 6 gives:
$$ x^2+3x-10=0.$$
We factor the quadratic equation to obtain:
$$(x+5)(x-2)=0.$$
Setting each factor equal to $0$ and solving for $x$, we get that $x=-5$ or $x=2$.